Simple & Compound Interest – Study Notes
(Jammu & Kashmir PSC – Quantitative Aptitude)
1. Simple Explanation
| Concept | What it is | When it is used |
|---|---|---|
| Simple Interest (SI) | Interest that is calculated only on the initial principal. | Projects with a fixed rate and a single compounding period (e.g., short‑term loans). |
| Compound Interest (CI) | Interest that is calculated on the principal + accumulated interest. | Long‑term investments, savings accounts, loans with periodic compounding. |
Both are expressed in terms of rate of interest per annum (r), time (t), and principal (P).
2. Key Formulae
| Type | Formula | Notes |
|---|---|---|
| Simple Interest | ( \displaystyle SI = \frac{P \cdot r \cdot t}{100} ) | r in % per annum, t in years. |
| Simple Interest + Principal | ( \displaystyle A = P + SI = P \left(1 + \frac{r t}{100}\right) ) | A = amount after t years. |
| Compound Interest (discrete) | ( \displaystyle A = P \left(1 + \frac{r}{n}\right)^{nt} ) | n = compounding periods per year (1 = yearly, 4 = quarterly, etc.) |
| Compound Interest (continuous) | ( \displaystyle A = P e^{rt/100} ) | e ≈ 2.71828; used rarely in exam problems. |
| Interest earned (CI) | ( \displaystyle CI = A - P ) | Amount of interest earned over t years. |
| Effective Annual Rate (EAR) | ( \displaystyle EAR = \left(1+\frac{r}{n}\right)^{n} - 1 ) | Useful for comparing different compounding frequencies. |
3. Quick Revision Points
| Topic | Revision Tip |
|---|---|
| SI | Remember “Principal × Rate × Time / 100”. |
| CI | Remember “Principal × (1 + Rate/n)^(n×Time)”. |
| Compounding Frequency | 1 = yearly, 2 = half‑yearly, 4 = quarterly, 12 = monthly, 365 = daily. |
| Converting Rates | If r is per month, multiply by 12 to get annual rate. |
| Working Backwards | Use logs to compute time: ( t = \frac{\ln(A/P)}{n \ln(1+r/n)} ). |
| Multiple Rates | Break the period into segments, apply formula for each segment. |
| Comparing Loans | Use the present value of payments to decide which loan is cheaper. |
4. Worked Examples
4.1 Simple Interest
Problem:
A person deposits ₹50,000 in a bank for 3 years at an annual simple interest rate of 6%. What is the amount at the end of 3 years?
Solution:
-
Identify variables:
- ( P = ₹50,000 )
- ( r = 6% ) per annum
- ( t = 3 ) years
-
Compute SI:
[ SI = \frac{P \cdot r \cdot t}{100} = \frac{50{,}000 \times 6 \times 3}{100} = \frac{900{,}000}{100} = ₹9,000 ] -
Compute final amount ( A ):
[ A = P + SI = 50{,}000 + 9{,}000 = ₹59{,}000 ]
Answer: ₹59,000.
4.2 Compound Interest (Quarterly)
Problem:
₹25,000 is invested at an annual rate of 8% compounded quarterly for 5 years. Find the amount at the end of 5 years.
Solution:
-
Variables:
- ( P = ₹25{,}000 )
- ( r = 8% )
- ( n = 4 ) (quarterly)
- ( t = 5 ) years
-
Apply CI formula:
[ A = P \left(1 + \frac{r}{n}\right)^{nt} = 25{,}000 \left(1 + \frac{8}{4}\right)^{4 \times 5} = 25{,}000 \left(1 + 2\right)^{20} = 25{,}000 \times 3^{20} ](Compute (3^{20}) ≈ 3 486 784.401)
[ A \approx 25{,}000 \times 3{,}486{,}784.401 = ₹87{,}169{,}610.025 ]
(Rounded to nearest rupee: ₹87,169,610)
-
Interest earned:
[ CI = A - P = 87{,}169{,}610 - 25{,}000 = ₹87{,}144{,}610 ]
Answer: ₹87,169,610.
4.3 Compound Interest (Monthly, Different Rates)
Problem:
₹10,000 is deposited at 5% per annum (compounded monthly) for 2 years. After that, the rate changes to 6% per annum (still